The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to:

The product of the slopes of the common tangents to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ and the parabola $y^2 = 8x$ is:

Let $F_1(-1, 0)$ and $F_2(1, 0)$ be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having its vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
$(1)$ The orthocentre of the triangle $F_1 M N$ is
$(A)$ $\left(-\frac{9}{10}, 0\right)$ $(B)$ $\left(\frac{2}{3}, 0\right)$ $(C)$ $\left(\frac{9}{10}, 0\right)$ $(D)$ $\left(\frac{2}{3}, \sqrt{6}\right)$
$(2)$ If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$,then the ratio of the area of the triangle $M Q R$ to the area of the quadrilateral $M F_1 N F_2$ is
$(A)$ $3: 4$ $(B)$ $4: 5$ $(C)$ $5: 8$ $(D)$ $2: 3$

Which one of the following is the common tangent to the ellipses $\frac{x^2}{a^2 + b^2} + \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} + \frac{y^2}{a^2 + b^2} = 1$?

Tangents are drawn from the point $P(3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,touching the ellipse at points $A$ and $B$. The orthocenter of $\Delta PAB$ is:

Find the equation of one of the common tangents to the parabola $y^2 = 4x$ and the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$.